APS22-4 2-Prop Z Procedures - Kelly Andrews Denney | Library

#1-8 Use the prompt below. 
One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later, it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him. 

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One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later, it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him.

Do these results indicate a significant decrease in voter support for his candidacy? What hypotheses would you test? p represents the true proportion of voters planning to vote for this candidate. b=before scandal, a=after scandal

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One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later, it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him.

What is the pooled (or combined) proportion? Use 340 out of 630 for before and 515 out of 1010 for after.

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One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later, it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him.

Do these results indicate a significant decrease in voter support for his candidacy? What is the mean for the model we're testing?

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One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later, it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him.

Do these results indicate a significant decrease in voter support for his candidacy? What calculations will find our standard deviation?

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One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later, it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him.

Do these results indicate a significant decrease in voter support for his candidacy? Calculate the z-score.

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One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later, it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him.

Do these results indicate a significant decrease in voter support for his candidacy? Calculate the p-value.

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One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later, it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him.

At a significance level of 0.05, do we have evidence of a significant decrease in voter support?

Below is a video of the full 2-prop z-test procedure. It shows how to pool/combine your proportions and how to calculate the standard deviation of the difference. Showing that work isn't strictly necessary, but I wanted you to see the process. There is also a small mistake in how I labeled the standard deviation. To see that proper notation, look at the formula sheet and how I showed it when I'm explaining how the book shows it.

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One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. A week later, it became known that he had had an extramarital affair, and a new poll showed only 51% of 1010 voters supporting him.

In reality, there was a significant drop in voter support for this candidate. Was an error made? If so, what kind?

For #9-14, use the prompt below.
Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed-killing herbicides on house pets. They examined 827 dogs from homes where an herbicide was used on a regular basis, diagnosing malignant lymphoma in 473 of them. Of the 130 dogs from homes where not herbicides were used, only 19 were found to have lymphoma.

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Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed-killing herbicides on house pets. They examined 827 dogs from homes where an herbicide was used on a regular basis, diagnosing malignant lymphoma in 473 of them. Of the 130 dogs from homes where not herbicides were used, only 19 were found to have lymphoma.

Is this 1-proportion or 2-proportion?

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Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed-killing herbicides on house pets. They examined 827 dogs from homes where an herbicide was used on a regular basis, diagnosing malignant lymphoma in 473 of them. Of the 130 dogs from homes where not herbicides were used, only 19 were found to have lymphoma.

Have the conditions for inference been met? (choose the most complete answer)

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Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed-killing herbicides on house pets. They examined 827 dogs from homes where an herbicide was used on a regular basis, diagnosing malignant lymphoma in 473 of them. Of the 130 dogs from homes where not herbicides were used, only 19 were found to have lymphoma.

Construct a 95% confidence interval for the difference in the two rates of lymphoma (herbicide-no herbicide). Enter you interval rounding each number to the nearest thousanth (three decimal places) with parentheses around your interval, a comma between, and no spaces. For example; (.123,.567)

The conclusion for a 2-sample interval can be really tricky. For a 2-prop z-interval, try to stick with the following:
I'm ____% confident the true proportion of (context here for group 1) is between ___% and ___%  (higher or lower) than (group 2).
If you have a negative and positive values in your interval, try this:
I'm ____% confident the true proportion of (context here for group 1) is between ___% lower and ___% higher than (group 2).
Don't feel bad if your conclusion feels clunky and weird. It's difficult to word these smoothly. I almost never do.

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Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed-killing herbicides on house pets. They examined 827 dogs from homes where an herbicide was used on a regular basis, diagnosing malignant lymphoma in 473 of them. Of the 130 dogs from homes where not herbicides were used, only 19 were found to have lymphoma.

Write the conclusion for the confidence interval. Remember to include context! This won't auto-grade well, so just check your answer against mine carefully, and let me know if you want me to verify whether your answer is correct.

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If there is no true difference between two independent groups, then the difference in the proportions would be ___. In symbols:

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What does your interval suggest about the connection between the use of herbicides and the incidence of lymphoma in dogs?

For the next few questions, use the prompt below.
The Centers for Disease Control and Prevention reported a survey of randomly selected Americans 65 and older, which found that 411 of 1012 men (group 1) and 535 of 1062 women (group 2) suffered from some form of arthritis.

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The Centers for Disease Control and Prevention reported a survey of randomly selected Americans 65 and older, which found that 411 of 1012 men (group 1) and 535 of 1062 women (group 2) suffered from some form of arthritis.

You may assume conditions for inference have been met. Use your calculator to find a 95% confidence interval for the difference in the proportion of men and women who have the disease.

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The Centers for Disease Control and Prevention reported a survey of randomly selected Americans 65 and older, which found that 411 of 1012 men (group 1) and 535 of 1062 women (group 2) suffered from some form of arthritis.

Write your conclusion. (again, autograde will almost certainly tell you this is wrong)

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The Centers for Disease Control and Prevention reported a survey of randomly selected Americans 65 and older, which found that 411 of 1012 men (group 1) and 535 of 1062 women (group 2) suffered from some form of arthritis.

Does your confidence interval provide evidence of a significant difference in the proportions of men and women who suffer from some form of arthritis?